Problem: $\dfrac{ -8r + 6s }{ 3 } = \dfrac{ -7r - 7t }{ -6 }$ Solve for $r$.
Solution: Multiply both sides by the left denominator. $\dfrac{ -8r + 6s }{ {3} } = \dfrac{ -7r - 7t }{ -6 }$ ${3} \cdot \dfrac{ -8r + 6s }{ {3} } = {3} \cdot \dfrac{ -7r - 7t }{ -6 }$ $-8r + 6s = {3} \cdot \dfrac { -7r - 7t }{ -6 }$ Multiply both sides by the right denominator. $-8r + 6s = 3 \cdot \dfrac{ -7r - 7t }{ -{6} }$ $-{6} \cdot \left( -8r + 6s \right) = -{6} \cdot 3 \cdot \dfrac{ -7r - 7t }{ -{6} }$ $-{6} \cdot \left( -8r + 6s \right) = 3 \cdot \left( -7r - 7t \right)$ Distribute both sides $-{6} \cdot \left( -8r + 6s \right) = {3} \cdot \left( -7r - 7t \right)$ ${48}r - {36}s = -{21}r - {21}t$ Combine $r$ terms on the left. ${48r} - 36s = -{21r} - 21t$ ${69r} - 36s = -21t$ Move the $s$ term to the right. $69r - {36s} = -21t$ $69r = -21t + {36s}$ Isolate $r$ by dividing both sides by its coefficient. ${69}r = -21t + 36s$ $r = \dfrac{ -21t + 36s }{ {69} }$ All of these terms are divisible by $3$ $r = \dfrac{ -{7}t + {12}s }{ {23} }$